# Rotationally invariant metrics and conservation of angular momentum

This was prompted by an exam question, though the questions are more general:

A 2D Riemannian space has the metric:

$$ds^2=dr^2 + \gamma^2 r^4 d\phi^2$$

State what conserved quantity corresponds to the independence of the metric coefficients on the angle $$\phi$$ , and show explicitly that it is ondeed conserved

Now I would have thought that this conserved quantity is angular momentum. However the angular geodesic equation using the Lagrangian $$L = g_{\mu\nu} \dot{x}^\mu\dot{x}^\nu$$ is $$r^4\dot\phi = constant$$ and not $$r^2\dot\phi = constant$$, which I would expect to corresond to the angular momentum. I tried to morph it into this form using the radial geodesic equation, but it doesn't work.

I realised that if the metric coefficients were any arbitrary function of $$r$$, I would have still said that anugular momnetum is the conserved quantity, and yet clearly we can choose functions of r so that it is impossible to get $$r^2\dot\phi$$ as the conserved quantity!

What is going on here? How can a rotationally invariant metric not correspond to a physical situation in which the angular momentum is conserved along geodesics?

• Define a new radial coordinate $\rho = \gamma r^2$ to put the metric into the form $ds^2=(d\rho)^2/(4\gamma^2\rho)+\rho^2\,d\phi^2$, which is fine except for an unimportant coordinate singularity at $\rho=0$. Now the conserved quantity has the familiar form $\rho^2\dot\phi$. With this in mind, how are you defining "angular momentum"? – Chiral Anomaly May 18 at 22:58
• @ChiralAnomaly Oh! So I suppose the discrepancy is whether one defines angular momentum as 'that quantity which is conserved from the roational invariance of our lagrangian' or as the familiar classical quanitiy L=mr^2\dot\phi. And in either case it is not entirely clear what the physical meaning of the quantities appearing in the lagrangian given are... – 21joanna12 May 19 at 13:34
• Right. The rotation-invariance-based definition of angular momentum is the more appropriate one because it's conserved, no matter how unfamiliar it may look in a given coordinate system. The result you derived is correct, even though it looks unfamiliar. – Chiral Anomaly May 19 at 20:04